I have some difficulties understanding something. There are several discretization methods, such as zero-order-hold (ZOH), forward euler, backward euler, tustin, et cetera.
- Forward euler, backward euler, et cetera discretization methods approximate the computation of a integral (see below), but what is the integral approximation when using a ZOH? What does a ZOH do?
- Why does Matlab not support forward euler, backward euler, Simpsons rule or even higher order integral approximations as discretization methods? But only ZOH, tustin, zero-pole matching? Is there a reason for it?
- Is it even usefull to use for instance Simpsons rule as a discretization method or even use higher order approximations? As far as I read, from http://en.wikibooks.org/wiki/Control_Systems/Z_Transform_Mappings#Simpson.27s_Rule, Simpsons rule has only one disadvantage?
For instance,
Consider the transfer function
$$\frac{F(s)}{E(s)} = \frac{1}{s}$$
This corresponds to the differential equation
$$\frac{\mathrm{d}f(t)}{\mathrm{d}t} = e(t)$$
Integrating both sides gives
$$f(t) = f(t_0) + \int_{t_0}^{t} e(t) \mathrm{d}t$$
Now t is evenly spaced, e.g., t = kT, with k = 0,1,2,... During one sampling t0 = kT and t = kT + T, the solution becomes
$$f(kT + T) = f(kT) + \int_{kT}^{kT + T} e(t) \mathrm{d}t$$
Now using the trapeziodal rule (tustin) http://en.wikipedia.org/wiki/Trapezoidal_rule. The integral is approximated by
$$ \int_{a}^{b} f(x)\, dx \approx (b-a) \left[\frac{f(a) + f(b)}{2} \right]$$
As a result, we obtain
$$f(kT + T) = f(kT) + \frac{kT + T - kT}{2} \left( e(kT) + e(kT + T) \right)$$
Using the z-transform you get
$$(z - 1)F(z) = \frac{T}{2} (z + 1) E(z) \rightarrow \frac{F(z)}{E(z)} = \frac{T}{2} \frac{z + 1}{z - 1}$$
and as such you determine by comparing the result with the very first equation that
$$s = \frac{2}{T} \frac{z - 1}{z + 1}$$
The same you can determine using forward euler, backward euler, et cetera. Now I am wondering what is the replacement for the laplace variable s when you will be using ZOH? And what kind of approximation does the ZOH use? As far as I understand you discretize the system using ZOH by applying
$$G(z) = (1 - z^{-1}) \mathcal{Z}\left\{\frac{G(s)}{s}\right\}$$